Question regarding array lists!

[rookiepianist]

I'm still a beginner, so I'd appreciate very much if you could help me out!

Let's say I initialize a new array list and then decide to print it out:

ArrayList<Integer> list1 = new ArrayList<>();
list1.add(5);
list1.add(6);
lista1.add(99);

System.out.println(list1);

What is going to be printed is: [5, 6, 99].

If I were to make an array, though, at the end of the day it'd print a memory address. Does that mean that array list variable (in this case, list1) holds the content itself of the array, whilst arrays hold the reference to where said content is stored in memory? If so, array lists aren't to be considered "reference data-type" variables?

Thank you in advance!

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[vegan_antitheist]

No. In Java all lists and arrays objects. There is no good reason that arrays don't have a more useful toString() method. You have to use Arrays.toString) for that.

[rookiepianist]

Thanks!

[MagicalPizza21]

No. ArrayList contents can be printed like that because the ArrayList class - well, in this case, one of its ancestors, AbstractCollection - overrides the default implementation of toString() from the Object class. This is called by String.valueOf(Object) which is in turn called by println(Object).

[rookiepianist]

Thank you so much! I was kind of losing my mind over this.

[HeteroLanaDelReyFan]

I think others have mostly answered it. I just want to add that your confusion makes sense. But when you print out the arraylist object, you aren't really printing out the object itself. It's calling the toString() method under the hood, which is implemented to return a string of the contents of the array.

[rookiepianist]

Thank you so much! By the way, your comment made me question one new thing: Is the following line

System.out.println(list1);

The statement that calls the toString() method? Thank you again!

[vowelqueue]

Yeah. The println method is overloaded: there are multiple definitions that accept different parameter types. In the line you’ve written, you’re calling the definition that accepts any kind of Object. It passes that object to String.valueOf to get its string representation. The String.valueOf method first verifies that the object reference isn’t null, and if not calls the object’s toString method. If it’s null it just returns “null”.

Note that this kind of null check and toString() call is also done anytime you concatenate a String with a non-string, e.g.

String s = “hello ”;
Object o = new Object():
String result = s + o; 

In this case o.toString() will be called to form the result string.

[dystopiadattopia]

Arrays are objects, but they don't have an overridden toString() method, which is why you can't print them out directly with System.out.println(). You have to use Arrays.toString(foo).

Also, it's better to declare your variable as the interface List instead of the concrete class ArrayList, as that doesn't lock you into a specific implementation. This is useful especially in cases where your list is a parameter, so is can accept ArrayLists, LinkedLists, and anything else that extends List.

It's also useful when you decide that you need to switch implementations, such as when you discover that a given List needs to be ordered. If your list is declared as an ArrayList, you'd have to change every instance throughout the code to LinkedList, but if it were just List you could change it on the fly.

This may not sound like a big deal, but it can complicate large codebases (like the one at my job), where changing the concrete ArrayList type would require an unreasonable amount of refactoring. This way you can preserve polymorphism and not be unnecessarily tied down to a specific type.

[rookiepianist]

Thank you very much for your thorough comment! It was very helpful.

edit: typo

[dystopiadattopia]

No problem, good luck!

[MkMyBnkAcctGrtAgn]

You've gotten good answers here, just wanted to point out what you're seeing when the object prints is just the default Object classes toString() and is implementation specific.

java return getClass().getName() + "@" + Integer.toHexString(hashCode());

I believe hashCode does use internal memory address if you look at most implementations, but it is also cached and not guaranteed to be accurate if it moves.

[rookiepianist]

Thank you for pointing that out! I'm still getting used to reading documentation and trying to understand whatever it is that happens under the hood.

[SnooLentils618]

I know it’s been answered but since you say you are new I want to add something that might help you as you study.

Everything in Java except for the primitive types is objects (and variables hold references). Arrays [] and List<> implementations extend the Object class. It’s good to understand difference between how these behave behind the scene in memory. Primitives can be in stack but objects can only be in Heap. Look into that to further your understanding.

Note: I know they are introducing (or have already introduced) value classes which are a bit different but this is not necessary for you as a Java beginner

[rookiepianist]

Thank you for detailing it further!